Question

Find the approximate value of f (5.001), where f (x) = x³ − 7x² + 15.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ x=5 \\ x+∆x=5.001 \\ \Rightarrow ∆x=0.001 \\ f\left(x\right)=x^3-7x^2+15 \\ \Rightarrow y=f\left(x=3\right)=125-175+15=-35 \\ \mathrm{Now},y=f\left(x\right) \\ \Rightarrow \frac{dy}{dx}=3x^2-14x \\ \therefore dy=∆y=\frac{dy}{dx}dx=\left(3x^2-14x\right)\times 0.001=\left(75-70\right)\times 0.001=0.005 \\ \therefore f\left(5.001\right)=y+∆y=-35+0.005=-34.995 \end{gathered}$$